CPhill

(x + 4) / (x + 5)  =  (x - 5 + 7)  / (2x)

(x + 4) / ( x + 5)  = ( x +2) / (2x)      cross-multiply

(2x) ( x + 4) =  (x + 5) ( x + 2)

2x^2 + 8x =   x^2  + 7x + 10        rearrange as

x^2 + x   =  10        complete the square on x

x^2 + x + 1/4 =  10 + 1/4

(x + 1/2)^2 =  41/4       take both roots

x + 1/2  =  ±  √41 / 2

x =  -1/2 ±  √41 / 2

(1 - 2) + (3 - 4)  + .... + (2005- 2006) + (2007 -2008) + ( 2009 -2010)

We have    (2010 / 2)  =     1005  pairs each summing to -1

So

1005 (-1) =  - 1005

Well....just ask.....somebody might know the answer

Draw QC and PC

Since angle QRC  =35

Then angle CRP  also = 35 

Then angle QRP  = 70

And angle   QPR =   180  - 68 - 70 =  180  - 138  =   42°

And you are EXACTLY correct !!!!

 In triangle ABC let the angle bisectors be BY and CZ. Given AB=8, AY=6, and CY=3, find BZ.

In the other problem we found that BC = 4

So, again, by the angle bisector theorem

AC / BC = AZ / BZ

9 / 4  =  AZ / BZ

So....there  are 13 parts to AB

And BZ  =  4 of them

So

BZ =  (4/13)(AB)  =  (4/13)(8) =   32/ 13

CORRECTED  !!!

 In triangle ABC let the angle bisectors be BY and CZ. Given AB=8, AY=6, and CY=3, find BC.

Angle  Bisector Theorem

BC / CY = BA / AY

BC / 3 = 8 / 6

BC  =  3 * 8 /  6  =    24 / 6   =   4

Since angle AIB  =  106

Then    angle (A  / 2) +   angle( B /2)  + 106  =180

A/2 + B/2 =  180-106

A/2 + B/2  =  74

A + B =  148

So

C =  180   -  148  =  32°

The common ratio is  r  / 2

And       l common ratio  l   <  1

So

-1  < r/2 <  1

-2 < r < 2

f(2)  = 2^2 + 2sqrt (2)  = 4 + 2sqrt (2)

2f(2)  = 8 + 4sqrt (2)

f(8)  = 8^2 + 2sqrt (8) =     64  + 2*2sqrt (2)  =  64 + 4sqrt (2)

2f(2) - f(8)  =

(8 + 4sqrt (2)  )  -  ( 64 + 4sqrt(2) )  =    - 56

The common  ratio  =  (2/3) / (3/2)  =  (2/3) (2/3)  =   4/9

Sum  =

(3/2)  / ( 1 - 4/9)  =

(3/2) / ( 5/9)  =   (3/2) ( 9/5)    =   27  / 10